Saturday, August 7, 2010

Problem 494: Circular Sector, 90 Degrees, Semicircle, Chord, Parallel

Geometry Problem
Click the figure below to see the complete problem 494 about Circular Sector, 90 Degrees, Semicircle, Chord, Parallel.

Problem 494. Circular Sector, 90 Degrees, Semicircle, Chord, Parallel.
See also:
Complete Problem 494

Level: High School, SAT Prep, College geometry


  1. Line AB intersect circle O’ at M. M is midpoint of AB . Triangle AO’M is isosceles right triangle
    Quadrilateral ACBD is a parallelogram and diagonals CD and AB intersect at midpoint M of AB and CD.
    E is midpoint of AC (corresponding points of dilation transformation centered at A with dilation factor =1/2 ).
    Angle ADC=45 (Angle face 90 degrees arc AM)
    Consider circle O’ and point C . We have CE.CA=CM.CD .
    Since M and E are the midpoints of CD and CA so CA=CD and CM=CE
    triangle ACD is isosceles right triangle.
    Triangles ACM congruence with DCE ( case SAS)
    So x=DE=AM =O’A*SQRT(2)=2*SQRT(2)

  2. Just confirming Peter's excellent solution by analytical geometry: Let O be (0,0). So A:(0,4),B:(4,0). The two circles are: x^2+y^2=16 & x^2+(y-2)^2=4. Now let C be (c,d) and D be (a,b). The given data can be translated into following eqns: a^+(b-2)^2=4, c^2+d^2=16,bc=(a-4)(d-4) and c^2+(d-4)^2=(a-4)^2+b^2 which when solved give us: a=8/5,b=4/5,c=12/5 and d=16/5.
    Now we can write the eqn. of AC as y=-x/3+4 and determine E as (6/5,18/5) as the intersection of the smaller circle and AC. Now x^2=(8/5-6/5)^2+(4/5-18/5)^2 or x^2=8 -->x=2√2

  3. ang AOC = ang AO'E (isoc tr & A common)
    ang COB = ang DO'O ( OD bisect)
    => AO'E + DO'O = 90° => EO'D = 90°
    => x² = 2² + 2²
    x = 2√2

  4. To c.t.o.e.:
    Will u pl explain what u mean by "ang COB=ang DO'O (OD bisect)". Unfortunately, it isn't very clear how u arrive at EO'D = 90° which, of course, has to be correct.

  5. To Vihaan:
    OD perpendicular to AD and CD (AD // CD)
    BOC isoceles
    => OD bisector

    DO'O = arc OD
    BOD = 1/2 arc OD ( OB tg to O' and OD chord )
    BOC = 2 BOD = arc OD
    => DO'O = BOC

    ( nice solution of 495 )

  6. Thanks. BTW, I'm very curious to know what "c.t.e.o" stands for!

  7. 1. Angle AEO is 90 degrees since AEO is an inscribed triangle on the diameter of the smaller circle.
    2. Angle ACB is 135 degrees since it inscribes the 270 degree arc on the larger circle.
    3. ACBD is a parallelogram by the problem definition and how it was constructed. So angle CAD is supplementary to ACB and 45 degrees.
    4. Let F be the intersection of AD and OE. Triangle AEF is a right isosceles by angle chasing. AE = EF and AF = sqrt(2) AE
    5. AEDO is a cyclic quadrilateral by definition on the smaller circle so AFO is similar to EDF and DE/ AO = EF / AF or DE = AO * EF / AF = 4 * AE / sqrt(2) * AE =
    2 sqrt(2)

    Bonus: AE = EC since ACO is isosceles. Further drawing in OC and doing some angle chasing shows CDO is congruent to OBD by SAS so CD = BD and triangle ACD and BCD are both isosceles right triangles. That means CD when extended intersects the other end of the diameter of the larger circle and in fact is 2x scaled version of AEO.

  8. Since <AOB = 90, <ACB = 135.
    Hence in \\ogram ADBC, <EAD = 45
    So <EO'D = 2 X 45 = 90

    So in isosceles Tr.DEO', x^2 = 2^2 + 2^2 = 8 and so x = 2sqrt2

    Sumith Peiris
    Sri Lanka

  9. Sumith

    Excellent solution !

    Peter Tran